Reactivity 3.1.11 — Conjugate Acid–Base Equilibrium Relationships

• Weak acids and bases exist in dynamic equilibrium with their ionized forms in aqueous solutions.
• Every acid has a conjugate base, and every base has a conjugate acid.

• The acid dissociation constant (\( K_a \)) and the base dissociation constant (\( K_b \)) are linked for any conjugate acid–base pair.
• This relationship arises from the ionic product of water \( (K_w) \), which is the equilibrium constant for the autoionization of water:
  \( \text{H}_2\text{O (l)} \rightleftharpoons \text{H}^+ (aq) + \text{OH}^- (aq) \)
• At 25°C, \( K_w = 1.0 \times 10^{-14} \).

Deriving the Relationship \( K_a \times K_b = K_w \):

• Consider a weak acid \( \text{HA} \) and its conjugate base \( \text{A}^- \).
• The dissociation of the acid is given by:
  \( \text{HA} (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{H}_3\text{O}^+ (aq) + \text{A}^- (aq) \)
  \( K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} \)
• The dissociation of the conjugate base is:
  \( \text{A}^- (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{OH}^- (aq) + \text{HA} (aq) \)
  \( K_b = \dfrac{[\text{OH}^-][\text{HA}]}{[\text{A}^-]} \)
• Multiplying the two expressions:
  \( K_a \times K_b = \left( \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} \right) \times \left( \dfrac{[\text{OH}^-][\text{HA}]}{[\text{A}^-]} \right) = [\text{H}_3\text{O}^+][\text{OH}^-] \)
• Therefore, \( K_a \times K_b = K_w \), since \( [\text{H}_3\text{O}^+][\text{OH}^-] = K_w \).

Key Takeaways:

• This equation allows conversion between \( K_a \) and \( K_b \) for a conjugate pair.
• If one value is known, the other can be calculated easily using:
  \( K_b = \dfrac{K_w}{K_a} \quad \text{or} \quad K_a = \dfrac{K_w}{K_b} \)
• It is valid only for conjugate acid–base pairs in aqueous solutions and assumes standard temperature (25°C) unless stated otherwise.

Why This Relationship Matters:

• Helps predict the behavior of weak acids or bases by linking their ionization constants.
• Useful in determining the relative strengths of conjugate pairs.
• Forms the mathematical foundation for converting between pKa and pKb:
  Since \( K_a \times K_b = K_w \), taking logarithms:
  \( \text{p}K_a + \text{p}K_b = \text{p}K_w = 14 \) (at 25°C)

Applying the Relationship Between \( K_a \), \( K_b \), and \( K_w \)

Step-by-Step Procedures:

• To calculate Kb from Ka:
  • Use the equation: \( K_b = \dfrac{K_w}{K_a} \)
  • Use \( K_w = 1.0 \times 10^{-14} \) at 25°C
• To calculate Ka from Kb:
  • Use the reciprocal: \( K_a = \dfrac{K_w}{K_b} \)
• To convert Ka to pKa:
  • Use \( \text{p}K_a = -\log_{10}(K_a) \)
• To convert pKa to Ka:
  • Use \( K_a = 10^{-\text{p}K_a} \)
• To convert between pKa and pKb:
  • Use: \( \text{p}K_b = 14 – \text{p}K_a \) or \( \text{p}K_a = 14 – \text{p}K_b \)

Important Notes for Exam Use:

• You are not expected to use the quadratic formula in any of these problems — only substitution and logarithmic manipulation.
• Values such as pKa or Ka are often given in the IB Chemistry data booklet or in the question itself.
• If a question gives pKa, you may need to convert it to Ka before using it in further calculations like equilibrium concentration or pH (in HL).
• Make sure the acid and base in question are conjugates before applying the \( K_a \times K_b = K_w \) relationship.

Application Examples:

• Calculating Kb for the conjugate base of a weak acid when Ka is given.
• Finding pKa of a weak acid when pKb of its conjugate base is known.
• Using Ka and Kb data to compare conjugate strengths and predict equilibrium direction.